State space representation See also
time-domain technique
x ˙ = A x + B u y = C x + D u \begin{align}
\dot{x} &= Ax + Bu \\\
y &= Cx + Du
\end{align} x ˙ y = A x + B u = C x + D u
independent State vector : x = [ x 1 , x 2 , … , x n ] T x = [x_{1},x_{2},\ldots, x_{n}]^{T} x = [ x 1 , x 2 , … , x n ] T
transfer function to a state space representation Given
G ( s ) = ∑ i = 1 n − 1 b i s i + b 0 s n + ∑ i = 1 n − 1 a i s i + a 0 = Y ( s ) U ( s ) G(s) = \frac{\sum_{i=1}^{n-1}b_is^i + b_{0}}{s^n + \sum_{i=1}^{n-1}a_is^{i} + a_{0}} = \frac{Y(s)}{U(s)} G ( s ) = s n + ∑ i = 1 n − 1 a i s i + a 0 ∑ i = 1 n − 1 b i s i + b 0 = U ( s ) Y ( s ) We get controller canonical state space form:
x ˙ ( t ) = [ 0 1 0 ⋯ 0 0 0 0 1 ⋯ 0 0 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 0 ⋯ 1 0 0 0 0 ⋯ 0 1 − a 0 − a 1 − a 2 ⋯ − a n − 2 − a n − 1 ] x ( t ) + [ 0 0 ⋮ 0 0 1 ] u ( t ) y ( t ) = [ b 0 b 1 ⋯ b n − 2 b n − 1 ] x ( t ) . \begin{aligned}
\dot{x}(t) &= \begin{bmatrix}
0 & 1 & 0 & \cdots & 0 & 0 \\\
0 & 0 & 1 & \cdots & 0 & 0 \\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\\
0 & 0 & 0 & \cdots & 1 & 0 \\\
0 & 0 & 0 & \cdots & 0 & 1 \\\
-a_0 & -a_1 & -a_2 & \cdots & -a_{n-2} & -a_{n-1}
\end{bmatrix} x(t) + \begin{bmatrix}
0 \\\
0 \\\
\vdots \\\
0 \\\
0 \\\
1
\end{bmatrix} u(t) \\\
y(t) &= \begin{bmatrix}
b_0 & b_1 & \cdots & b_{n-2} & b_{n-1}
\end{bmatrix} x(t).
\end{aligned} x ˙ ( t ) y ( t ) = 0 0 ⋮ 0 0 − a 0 1 0 ⋮ 0 0 − a 1 0 1 ⋮ 0 0 − a 2 ⋯ ⋯ ⋱ ⋯ ⋯ ⋯ 0 0 ⋮ 1 0 − a n − 2 0 0 ⋮ 0 1 − a n − 1 x ( t ) + 0 0 ⋮ 0 0 1 u ( t ) = [ b 0 b 1 ⋯ b n − 2 b n − 1 ] x ( t ) . We get observer canonical state space form:
x ˙ ( t ) = [ − a n − 1 1 0 ⋯ 0 0 − a n − 2 0 1 ⋯ 0 0 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ − a 2 0 0 ⋯ 1 0 − a 1 0 0 ⋯ 0 1 − a 0 0 0 ⋯ 0 0 ] x ( t ) + [ b n − 1 b n − 2 ⋮ b 2 b 1 b 0 ] u ( t ) y ( t ) = [ 1 0 ⋯ 0 0 ] x ( t ) . \begin{aligned} \dot{x}(t) &= \begin{bmatrix} -a_{n-1} & 1 & 0 & \cdots & 0 & 0 \\ -a_{n-2} & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -a_2 & 0 & 0 & \cdots & 1 & 0 \\ -a_1 & 0 & 0 & \cdots & 0 & 1 \\ -a_0 & 0 & 0 & \cdots & 0 & 0 \end{bmatrix} x(t) + \begin{bmatrix} b_{n-1} \\ b_{n-2} \\ \vdots \\ b_2 \\ b_1 \\ b_0 \end{bmatrix} u(t) \\ y(t) &= \begin{bmatrix} 1 & 0 & \cdots & 0 & 0 \end{bmatrix} x(t). \end{aligned} x ˙ ( t ) y ( t ) = − a n − 1 − a n − 2 ⋮ − a 2 − a 1 − a 0 1 0 ⋮ 0 0 0 0 1 ⋮ 0 0 0 ⋯ ⋯ ⋱ ⋯ ⋯ ⋯ 0 0 ⋮ 1 0 0 0 0 ⋮ 0 1 0 x ( t ) + b n − 1 b n − 2 ⋮ b 2 b 1 b 0 u ( t ) = [ 1 0 ⋯ 0 0 ] x ( t ) .